Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $20.5$ years; the standard deviation is $2.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living longer than $18$ years.
Solution: $20.5$ $18$ $23$ $15.5$ $25.5$ $13$ $28$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $20.5$ years. We know the standard deviation is $2.5$ years, so one standard deviation below the mean is $18$ years and one standard deviation above the mean is $23$ years. Two standard deviations below the mean is $15.5$ years and two standard deviations above the mean is $25.5$ years. Three standard deviations below the mean is $13$ years and three standard deviations above the mean is $28$ years. We are interested in the probability of a tiger living longer than $18$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $18$ years and the other half $({16\%})$ will live longer than $23$ years. The probability of a particular tiger living longer than $18$ years is ${68\%} + {16\%}$, or $84\%$.